Find the derivative of $\frac{(2x−1)e^{−2x}}{(1−x)^2}$

Question

I need to find the derivative of $$\frac{(2x−1)e^{−2x}}{(1−x)^2}$$

I seems very complex to me so I'm wondering if there is a rule or formula I should be using? I attempted it using the chain rule first for the numerator (since I have $ ( 2 x- 1)$ multiplied by $e^{- 2 x}$ as my numerator) and then my plan was to use this rule: $(\frac{u}{v})′=\frac{vu′-uv′}{v^2}$.

It gets messy and complicated. Could someone please explain how you'd attempt this problem?

I did write it properly. Someone ruined it. How do I stop people changing it? — Helena Shaw, Feb 27 '14 at 23:25
Hopefully people will let it alone: there's no mistaking the fraction, now! — amWhy, Feb 27 '14 at 23:29

score 3 · Accepted Answer · answered Feb 27 '14 at 23:29

3

If you want to differentiate the product of $3$ functions, the product rule is still available:

$$ (fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh' $$

(and if you had decided to group (gh) or (fh) at the outset, you can check you'll still get the same outcome)

So, if you take $f = 2x -1 $, $g = e^{-2x}$ and $h = (1-x)^{-2}$, you're set (notice I changed the exponent on $h$ to make this a product rule; you could similarly formulate things in terms of the quotient rule if you prefer)

answered Feb 27 '14 at 23:29

BaronVT

13,613

It's a quotient, not a product. Someone kept undoing the fraction that was originally posted. Of course, one could write the given quotient as a product, as well! – amWhy Feb 27 '14 at 23:30
hah, ok. I have edited my answer to account for this. – BaronVT Feb 27 '14 at 23:32
1

Thanks BaronVT and amWhy. That's exactly what I'm looking for. – Helena Shaw Feb 27 '14 at 23:34

mahavir · Answer 2 · 2014-02-28T05:14:33.120

1

$$\frac{d}{dx}\frac{(2x−1)e^{−2x}}{(1−x)^2}=\frac{(1-x)^2\frac{d}{dx}[(2x−1)e^{−2x}]-(2x−1)e^{−2x}\frac{d}{dx}{(1−x)^2}}{(1-x)^4} $$ I hope now you can solve easily

edited Feb 28 '14 at 05:14

answered Feb 27 '14 at 23:40

mahavir

753

I think you have an error in the second term of the numerator of your fraction; you should be differentiating $(1-x)^2$ for that term. – BaronVT Feb 27 '14 at 23:53

score -2 · Answer 3 · answered Feb 27 '14 at 23:33

$$ \dfrac{\mathrm{d}}{\mathrm{d}x}f\left(x\right) = \dfrac{-2{\cdot}\left(2x-1\right){\cdot}{\mathrm{e}}^{-\left(2x\right)}}{{\left(1-x\right)}^{2}}+\dfrac{2{\cdot}\left(2x-1\right){\cdot}{\mathrm{e}}^{-\left(2x\right)}}{{\left(1-x\right)}^{3}}+\dfrac{2{\mathrm{e}}^{-\left(2x\right)}}{{\left(1-x\right)}^{2}} $$

Find the derivative of $\frac{(2x−1)e^{−2x}}{(1−x)^2}$

3 Answers3